In order to know what is going on you have to know basic calculus.

I will display the symbol pi as "pi" throughout this

Here's the problem:
Find the minimum of the function y= 3 + 4cos(x) + cos(2x) over the interval [0,2pi] and also show that it is a minimum using concavity.

Easy peasy right?
First find the critical points
y'= -4sin(x) - 2sin(2x)
y'= -4sin(x) - 4sin(x)cos(x) (by double angle identity)
y'= -4sin(x)(1 + cos(x))
sin(x) is zero at 0 and 2pi and (1 + cos(x)) is zero at pi
Critical Points are: 0, pi, 2pi

Find position of critical points

f(0)= 3 + 4cos(0) + cos(0)
f(0)= 3+4+1 = 8

f(pi)= 3 + 4cos(pi) + cos(2pi)
= 3 - 4 + 1 = 0

f(2pi)= 3 + 4cos(2pi) + cos(4pi)
3 + 4 + 1 = 8

Ok so when x = pi is the minimum, barring any breaks (which there aren't any)

Now for concavity.
y'= -4sin(x) - 2sin(2x)
y''= -4cos(x) - 4cos(2x)
f''(pi)= 4 - 4 = 0

Wait zero? That means it's a point of inflection, not a minimum... but it's the lowest point on the graph? Actually if you look at the graph, it is the bottom of a periodic function!

Therefore, math is broken and illogical.

Someone please explain this to me

Whenever I was in Maths class in school, we used to shout at the teacher "when in real life would we EVER need to know this, or work this out". I got an A, and he ended up having a mid-life crisis.

Post has been edited 1 time(s), last time on Oct 29 2011, 2:56 am by DevliN. Reason: Mmkay.

Lol.

cos(pi)= -1
cos(2pi)= 1
-4(-1) - (4)(1)
or 4 - 4
also known as zero.

No error there, as far as I see...

Post has been edited 1 time(s), last time on Oct 29 2011, 3:00 am by DevliN. Reason: Deleted the picture from Card's post.

It's the fact that you think a value of zero for the second derivative = point of inflection that's the problem here.

Fine, I didn't know that it was possible to have a second derivative of zero and still not be a point of inflection.

1. How is this possible, by the way. Explain it to me please.
2. For something to be a local minimum it HAS to be concave up, right? If not, why? Please explain this in great detail please.

It's possible because a POI isn't when the second derivative is zero, it's when the second derivative intersects the x-axis.

Is it not an intersection if it touches at one point?

You're saying there are always two points of inflection in a graph?

I was talking about this with a friend and we established that there isn't a point of inflection. I'm more concerned with the fact that it has a 2nd derivative of zero and is a minimum. How does that make sense?

No. y = x does not have concavity; that's the most basic example I can think of. But I can show you a a trigonometric example:



That's Sin[x], x = {[1, pi/2]}. The local minimum is clearly at Sin[1], which is concave down.

All this stuff reminds me that I've completely forgotten precalculus. Will I need to know any of this for calculus?

Yes.


No, that's a global minimum for the closed interval, it is not a local minimum.

Quote from name:Dem0nS1ayer

All this stuff reminds me that I've completely forgotten precalculus. Will I need to know any of this for calculus?

You can't get basic physics, so no matter how much precalc you have you won't get calculus. That point aside, you don't need this concavity bullshit for calculus. In fact, almost all of precalculus isn't used at all in calculus.

? How are you defining local and global minimum? ME NO UNDERSTAND.

The local extrema are critical points determined relative to the points on either side of them.

A global extremum is a maximum or minimum value of the entire interval being considered for the function.

And because the minima is on the boundary, it can't be approached from either side.

DT explained it all to me; I get it now.

Post has been edited 2 time(s), last time on Oct 29 2011, 5:56 am by Sacrieur.

I don't know how much of this has been explained, but

Your incorrect assumption is that a function has a local minimum at x = a if and only if f'(a) = 0 and f''(a) > 0. The latter condition is sufficient to show a local minimum, but not necessary. If the second derivative is also zero at a critical point, you do not have enough information to determine the local concavity or the type of critical point. (Second Derivative Test)

Consider the simple function f(x) = x⁴. This function has a critical point at x = 0, as well as a local minimum, but note that f'(0) = f''(0) = 0. The function is in a way "flatter", and you just don't know enough about its behavior at the point.

The terminator to all this.
lim(x->a+0)f'(x)*lim(x->a-0)f'(x) < 0 (f(x) is any function that is continuous. It however does not mean that f'(x) has to be continuous.)

If f'(x) is a derivative of function f, then x=a is a critical point for f.
If f'(x) is a second derivative of function F, then x=a is a inflection point for F.

Contrary to Azrael's and DTBK's nonsense, your incorrect assumption was to do fake math instead of real math. Any real mathematician would just plot the function and point at the lowest point, getting the answer first and not giving a shit about the rest. Then, 10 years later when the academy you published your answer to demands justification, you, annoyed that they don't believe your obvious work, go back and muddle through some nonsense about epsilons and alphas and send it back in, where a few decades later some poor graduate student finds it crumpled up next to a trash can, rewrites it neatly for you and then publishes it under his name. Fifty years later, someone who actually understands it will redo your proof in an easier more intuitive way, which you should then copy and send in to your teacher for credit.

Also, Azrael, a global minimum happens to also be a local minimum.

No, it isn't. It can be, but as shown by Sacrieur's example, it isn't always.